math lessons - Cardioid

In geometry, the cardioid is an epicycloid which has one and only one cusp. That is, a cardioid is a curve that can be produced as a locus — by tracing the path of a chosen point of a circle which rolls without slipping around another circle which is fixed but which has the same radius as the rolling circle.

The cardioid is also a special type of limaçon: it is the limaçon with one cusp.

The name comes from the heart shape of the curve (Greek kardioeides = kardia:heart + eidos:shape). Compared to the ♥ symbol, though, it doesn't have the sharp point at the bottom.

The cardioid is an inverse transform of a parabola.

The large, central, black figure in a Mandelbrot set is a cardioid. This cardioid is surrounded by a fractal arrangement of circles.

Contents

Equations

Since the cardioid is an epicycloid with one cusp, its parametric equations are

$x(\theta) = \cos \theta + {1 \over 2} \cos 2 \theta, \qquad \qquad (1)$

$y(\theta) = \sin \theta + {1 \over 2} \sin 2 \theta. \qquad \qquad (2)$

The same shape can be defined in polar coordinates by the equation

$\rho(\theta) = 1 + \cos \theta. \$

Proof

Equations (1) and (2) define a cardioid whose cuspidal point is (−1/2, 0). To convert to polar, the cusp should preferably be at the origin, so add 1/2 to the abscissa:

$x(\theta) = {1 \over 2} + \cos \theta + {1 \over 2} \cos 2 \theta,$

$y(\theta) = \sin \theta + {1 \over 2} \sin 2 \theta.$

The polar radius $ρ(θ)$ is given by

$\rho(\theta) = \sqrt{x^2(\theta) + y^2(\theta)}$

$= \sqrt{\left( {1 \over 2} + \cos \theta + {1 \over 2} \cos 2 \theta \right)^2 + \left( \sin \theta + {1 \over 2} \sin 2 \theta \right)^2 }.$

Expand,

$\rho = \sqrt{ {1 \over 4} + \cos^2 \theta + {1 \over 4} \cos^2 2 \theta + \cos \theta + {1 \over 2} \cos 2 \theta + \cos \theta \cos 2 \theta + \sin^2 \theta + {1 \over 4} \sin^2 2 \theta + \sin \theta \sin 2 \theta}.$

Simplify by noticing that

$\cos^2 \theta + \sin^2 \theta = 1, \qquad \qquad \mbox{(trig. ident.)}$

${1 \over 4} \cos^2 2 \theta + {1 \over 4} \sin^2 2 \theta = {1 \over 4}, \qquad \qquad \mbox{(variation of the above)}$

$\cos \theta \cos 2 \theta + \sin \theta \sin 2 \theta = \cos (\theta - 2 \theta) = \cos -\theta = \cos \theta. \$

Thus,

$\rho = \sqrt{ {1 \over 4} + 1 + {1 \over 4} + 2 \cos \theta + {1 \over 2} \cos 2 \theta }$

$= \sqrt{ {3 \over 2} + {4 \over 2} \cos \theta + {1 \over 2} \cos 2 \theta }$

$= \sqrt{ {3 + 4 \cos \theta + \cos 2 \theta \over 2}}.$

Then, since

$\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1, \qquad \qquad \mbox{(trigonometric identity)}$

it follows that

$\rho = \sqrt{ {3 + 4 \cos \theta + 2 \cos^2 \theta - 1 \over 2}} = \sqrt{ {2 + 4 \cos \theta + 2 \cos^2 \theta \over 2}},$

$\rho = \sqrt{ 1 + 2 \cos \theta + \cos^2 \theta} = 1 + \cos \theta,$

quod erat demonstrandum.

Graphs

Four graphs of cardioids oriented in the four cardinal directions, with their respective polar equations.

External link

Hearty munching on cardioids

Categories: Curves

01-04-2007 01:18:14
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Cardioid

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Mathematics Encyclopedia and Lessons

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Cardioid

Equations

Proof

Graphs

External link

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